Question 25165
{{{1/(x-1)+5=11/(x-1)}}}


There are a couple of ways to do this.  Probably best is to multiply both sides by the LCD which is (x-1).  This is valid, provided that x does NOT equal 1, because that would make the denominator zero, which is NEVER allowed!  


{{{(x-1) *(1/(x-1))+5*(x-1) =(x-1)*11/(x-1))}}}
{{{1 + 5x-5 = 11}}}
{{{5x -4 = 11}}}
{{{5x = 15}}}
{{{x=3}}}


Check: 
{{{1/(3-1) + 5 = 11/(3-1)}}}
{{{1/2 + 5 = 11/2}}}
True!  It checks!!


R^2 at SCC